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3 years ago

Which is the better buy, 7 pounds for $8.47 or 9 pounds for $11.07?

2 answers:

wel3 years ago

7 0

Let us take the case of 7 pounds for $8.47 first.
7 pounds of a product costs = 8.47 dollars
Then
1 pound of the same product will cost = (8.47/7) dollars
= 1.21 dollars
Now let us take the case of 9 pounds for $11.07
9 pounds of a product costs = 11.07 dollars
Then
1 pound of the same product will cost = (11.07/9) pounds
= 1.23 dollars
So from the above deductions we can see that 9 pounds for $11.07 is a better buy than 7 pounds for $8.47. I hope the procedure is clear enough for you to understand. In future you can use this method for solving similar problems.

Dennis_Churaev [7]3 years ago

7 0

7 for 8.47 because 8.47/7=1.21 and 11.07/9=1.23

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3 years ago

Binomial Expansion/Pascal's triangle. Please help with all of number 5.

Mandarinka [93]

$\begin{matrix}1\\1&1\\1&2&1\\1&3&3&1\\1&4&6&4&1\end{bmatrix}$

The rows add up to $1,2,4,8,16$ , respectively. (Notice they're all powers of 2)

The sum of the numbers in row $n$ is $2^{n-1}$ .

The last problem can be solved with the binomial theorem, but I'll assume you don't take that for granted. You can prove this claim by induction. When $n=1$ ,

$(1+x)^1=1+x=\dbinom10+\dbinom11x$

so the base case holds. Assume the claim holds for $n=k$ , so that

$(1+x)^k=\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k$

Use this to show that it holds for $n=k+1$ .

$(1+x)^{k+1}=(1+x)(1+x)^k$
$(1+x)^{k+1}=(1+x)\left(\dbinom k0+\dbinom k1x+\cdots+\dbinom k{k-1}x^{k-1}+\dbinom kkx^k\right)$
$(1+x)^{k+1}=1+\left(\dbinom k0+\dbinom k1\right)x+\left(\dbinom k1+\dbinom k2\right)x^2+\cdots+\left(\dbinom k{k-2}+\dbinom k{k-1}\right)x^{k-1}+\left(\dbinom k{k-1}+\dbinom kk\right)x^k+x^{k+1}$

Notice that

$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!}{\ell!(k-\ell)!}+\dfrac{k!}{(\ell+1)!(k-\ell-1)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)}{(\ell+1)!(k-\ell)!}+\dfrac{k!(k-\ell)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(\ell+1)+k!(k-\ell)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{k!(k+1)}{(\ell+1)!(k-\ell)!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dfrac{(k+1)!}{(\ell+1)!((k+1)-(\ell+1))!}$
$\dbinom k\ell+\dbinom k{\ell+1}=\dbinom{k+1}{\ell+1}$

So you can write the expansion for $n=k+1$ as

$(1+x)^{k+1}=1+\dbinom{k+1}1x+\dbinom{k+1}2x^2+\cdots+\dbinom{k+1}{k-1}x^{k-1}+\dbinom{k+1}kx^k+x^{k+1}$

and since $\dbinom{k+1}0=\dbinom{k+1}{k+1}=1$ , you have

$(1+x)^{k+1}=\dbinom{k+1}0+\dbinom{k+1}1x+\cdots+\dbinom{k+1}kx^k+\dbinom{k+1}{k+1}x^{k+1}$

and so the claim holds for $n=k+1$ , thus proving the claim overall that

$(1+x)^n=\dbinom n0+\dbinom n1x+\cdots+\dbinom n{n-1}x^{n-1}+\dbinom nnx^n$

Setting $x=1$ gives

$(1+1)^n=\dbinom n0+\dbinom n1+\cdots+\dbinom n{n-1}+\dbinom nn=2^n$

which agrees with the result obtained for part (c).

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3 years ago

7 + 2y = 8x 3x - 2y = 0 Solve the system of equations by substitution. (70/3, 140/9) (7/5, 21/10) no solution coincident

Wittaler [7]

So we have the system of equations:
$7+2y=8x$ equation (1)
$3x-2y=0$ equation (2)

To use substitution, we are going to solve for one variable in one of our equations, and then we are going to replace that value in the other equation:
Solving for $x$ in equation (2):
$3x-2y=0$
$3x=2y$
$x= \frac{2}{3}y$ equation (3)

Replacing equation (3) in equation (1):
$7+2y=8x$
$7+2y=8( \frac{2}{3} y)$
$7+2y= \frac{16}{3} y$
$7= \frac{10}{3} y$
$y= \frac{7}{ \frac{10}{3} }$
$y= \frac{21}{10}$ equation (4)

Replacing equation (4) in equation (3):
$x= \frac{2}{3}y$
$x=( \frac{2}{3} )( \frac{21}{10} )$
$x= \frac{7}{5}$

We can conclude that the solution of our system of equations is <span>(7/5, 21/10)</span>

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3 years ago

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ruslelena [56]

12:20

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3 years ago